The general solution of the differential equation is

Question

Question

This question was previously asked in

ESE Electrical 2017 Official Paper

Answer 1

Concept:

General equation for DE:

Then its corresponding Auxiliary equation will be

AE: Dⁿ + k₁ D^n-1 + … k_n = 0

Then the solution of above DE will be as follows:

Roots of Auxiliary Equation	Complementary Function
m_1,m₂, m₃, … (real and different roots)
m_1,m₁, m₃, … (two real and equal roots)
m_1,m₁, m₁, m₄… (three real and equal roots)
α + i β, α – i β, m₃, … (a pair of imaginary roots)
α ± i β, α ± i β, m₅, … (two pairs of equal imaginary roots)

Calculation:

Given:

Its Auxillary equations:

(D⁴ – 2D³ + 2D² – 2D + D) = 0

(D - 1)(D³ - D² + D - 1) = 0

(D - 1)(D - 1)(D2 + 1) = 0

It roots = 1, 1, i, -i

Then it has two equal roots and one pair of imaginary roots.

Therefore, the solution is:

(c₁ + c₂x) e^x + c₃ cos x + c₄ sin x = 0